10:29 ..."and that would all be fine, and we would be done! But that's not satisfying; it's not a nice, *clean* answer. So let's see if we can..." [sees there are over 7 minutes left in the video] [closes UA-cam...] [slowly backs away...]
This is the longest problem I've ever seen in my career as Math student & it requires many aspects of Trigonometry. I call this a do or die problem, there's no in between.
Sal - at 14:06 in your video when you want an expression for the cos(theta) use the geometric triangle representing angle theta where you already know the sin(theta) so opposite it (x-3) and hypotenuse is 2... You can find the cos(theta) using Pythagorean theorem so cos (theta) = (1/2)SQRT(4 - (x-3)^2).... reduces all the good algebra you are doing.
At, time 11:50, the formula for Sin(A+B)=SinA.CosB+SinA+CosB in place of Sin(A+B)=SinA.CosB+SinA.CosB=2SinA.CosB, A=B= theta. Otherwise, Great video. Didn't understand a thing in class. The video might be long but it make problems like this crystal clear. Thx
At 8:07, is it possible to factor out the 4 from the integration and make cos^2(theta) equal to 1 - sin^2(theta) and use the equations we already proved to solve the problem?
You can use u-substitution here. Let u = x - 3 Hence, du = dx integral of sqrt(4 - u) du which is just equal to arcsin(u/2) + c substitute back x - 3, you get arcin((x-3)/2) + c
He uses (sin(theta))^2 as the substitution because it satisfies the identity cos(theta) = 1 - sin(theta)^2. With this he can easily simplify the integral, as well as solve for x in terms of theta.
@tringuyen552911 Recall that to do d/dtheta (sin2(theta)) you must utilize the chain rule, so first you take the dervitative of the innermost portion, 2theta, which is 2, and then the derivative of the outer part, sin, which is cos, but carry the original inner through, so the derivative of sin2theta is 2cos2theta, which is what you originally have in the integral, so Sal did it right.
In working thru this, I let cos(theta) = (x-3)/2 instead of using sin(theta) as in the video. My answer differs from this one only in the arcsin term. I find my answer varies by a factor of pi/2 for this term. I cannot figure out why. Instead of 2arcsin((x-3)/2), I get -2arccos((x-3)/2). If I add a pi/2 to this, I can get to the correct answer. Can anyone give any help on this? Thanks!
In the Indian system of education we actually have a direct formula to use for problems of this type.This method however is really good to understand how we got there though.
This video really helped me. Thanks. But if you can, can you make a vid on this question.... the integral of (1/((4-x^2)^3/2)) dx. I have tried using the trig functions but never seem to get the right answer.
started working on this before watching the video- an hour later, i have 4 A4s filled, i am stuck in a loop of neverending integrals of sin(2x)thetas or cos(2x)dthetas lol. e: okay, so i forgot to apply the square identity after getting to integral of f*cos^2theta * dtheta. woops.
@@NOMVrewq It's a constant and in the squarroot it's +4 so it stays positive. You only have to be careful with + or - if you take x^2 out of the root, then it becomes abs(x).
dont understand why u first cancel those 2 from the ( x-3)/2 and then in antoehr line multiply with 2/2... and before u done that u could write sqrt( 1- (x-3)^2 /4) as sqrt(4 - ( x-3)^2)... ?
I think he messed up, the calculus is fine but it should be 1- (x-3/4)^2, not 1- (x-3/2)^2. I don't think you can take the 4 out of the square root like that.
He just used this identity: (a*b)^c = a^c*b^c In this case, a is 4, b is 1-(x-3)^2/4, and c is 1/2 Therefore, (4*(1-(x-3)^2/4)^(1/2) is equivalent to 4^(1/2) times (1-(x-3)^2/4)^(1/2) which is equivalent to 2(1-(x-3)^2/4)^(1/2) A square root is the same thing as taking the expression to the 1/2 power in case you didn't know. I know this reply is too late, but it might help other people that think like you.
10:29 ..."and that would all be fine, and we would be done! But that's not satisfying; it's not a nice, *clean* answer. So let's see if we can..."
[sees there are over 7 minutes left in the video]
[closes UA-cam...]
[slowly backs away...]
Awesome comment :D
9:52 +C has entered the chat.
15:12 +C has left the chat.
15:45 +C has entered the chat.
17:18 +C has left the chat, never to be seen again...
All that work and he forgot the constant at the end lol
This is the longest problem I've ever seen in my career as Math student & it requires many aspects of Trigonometry. I call this a do or die problem, there's no in between.
true
No +C? You get a C+...
Ahh man hate those teachers
HOW COULD FORGET +C?????
Sal - at 14:06 in your video when you want an expression for the cos(theta) use the geometric triangle representing angle theta where you already know the sin(theta) so opposite it (x-3) and hypotenuse is 2... You can find the cos(theta) using Pythagorean theorem so cos (theta) = (1/2)SQRT(4 - (x-3)^2).... reduces all the good algebra you are doing.
you forgot the + C
This was really a super problem on integration with trig substitution.
Thank you for the calculus videos I have viewed thus far. They have been helpful in building in intuition of the subject.
At, time 11:50, the formula for Sin(A+B)=SinA.CosB+SinA+CosB in place of Sin(A+B)=SinA.CosB+SinA.CosB=2SinA.CosB,
A=B= theta. Otherwise, Great video. Didn't understand a thing in class. The video might be long but it make problems like this crystal clear. Thx
Minus 1 point for no constant.....Did that once on a test and I will never do that again.
At 8:07, is it possible to factor out the 4 from the integration and make cos^2(theta) equal to 1 - sin^2(theta) and use the equations we already proved to solve the problem?
11:50 correct proof for the trig identity of sin(2x) is: sin(2x)=sin(x)cos(x)+cos(x)sin(x)=2sin(x)cos(x)
+C
You can use u-substitution here.
Let u = x - 3
Hence, du = dx
integral of sqrt(4 - u) du
which is just equal to arcsin(u/2) + c
substitute back x - 3, you get arcin((x-3)/2) + c
If it takes that long to do it, it's not gonna be on the exam, right?
lol. Good one.
And this is were you are WRONG!
@NOMVrewq It stays positive. You only add a plus minus sign when you get rid of the square root sign.
He uses (sin(theta))^2 as the substitution because it satisfies the identity
cos(theta) = 1 - sin(theta)^2. With this he can easily simplify the integral, as well as solve for x in terms of theta.
@tringuyen552911 Recall that to do d/dtheta (sin2(theta)) you must utilize the chain rule, so first you take the dervitative of the innermost portion, 2theta, which is 2, and then the derivative of the outer part, sin, which is cos, but carry the original inner through, so the derivative of sin2theta is 2cos2theta, which is what you originally have in the integral, so Sal did it right.
this got complicated when you tried to make it into a "nice clean answer">.< ...
In working thru this, I let cos(theta) = (x-3)/2 instead of using sin(theta) as in the video. My answer differs from this one only in the arcsin term. I find my answer varies by a factor of pi/2 for this term. I cannot figure out why. Instead of 2arcsin((x-3)/2), I get -2arccos((x-3)/2). If I add a pi/2 to this, I can get to the correct answer. Can anyone give any help on this? Thanks!
In the Indian system of education we actually have a direct formula to use for problems of this type.This method however is really good to understand how we got there though.
Can you share this formula, pls?
You are a miracle worker.
OHMAGAD I GET IT NOW! a week in university = 18 minutes on youtube.
Couldn't you use U-Substitution in this problem since x^2 and its derivative are technically present?
Thanks for the video. I also have some integration videos of my own on my channel.
It is a very good question!!!!
Thanks, helped me a lot.
This video really helped me. Thanks.
But if you can, can you make a vid on this question.... the integral of (1/((4-x^2)^3/2)) dx. I have tried using the trig functions but never seem to get the right answer.
started working on this before watching the video- an hour later, i have 4 A4s filled, i am stuck in a loop of neverending integrals of sin(2x)thetas or cos(2x)dthetas lol.
e: okay, so i forgot to apply the square identity after getting to integral of f*cos^2theta * dtheta. woops.
instead of doin all that after 13:15, you could just draw a right triangle with angle theta and use SOHCAHTOA to get the required angles
Beautiful 😍😍😍🔥🔥
You forgot plus C at the very end, but Brava! Brava!
how did you go from (x+3)^2 over 4 to (x+3)^2 over 2?? inside the root is subtraction so you cannot do that
I wish my teacher explained things this well...
LOL! He said "little" chalk board at the end.......
Yes... that was little to fit all of that.
sallu miya, great!
Can I substitute cosine square theta instead of sine square theta ?
Thank you sir.
I find it funny that his "simplification" cover a third to a half of the written work.
no problem this man khant solve
At 4:52, shoudn't it be (x-3)/2=+/- sin theta?
Great video! I just have 2 questions. Should the «2» become a «+ or - 2» when you pull it out from the square root at 3:35? Does it even matter?
It's +2.
@@quinten249 i know you are right but it seems almost arbitrary since it could have just as well been a -2... do you know why its +2? thanks!
@@NOMVrewq It's a constant and in the squarroot it's +4 so it stays positive. You only have to be careful with + or - if you take x^2 out of the root, then it becomes abs(x).
@@quinten249 sorry im not talking about the 1st 2, but the 2nd. he went from (x-3)^2/4 to (x-3/2)^2. im saying it could have just as been (x-3/-2)^2.
@@quinten249 I think i worded my original question incorrectly so I understand the confusion
amazing
OMG thank you SOooo much!!
dont understand why u first cancel those 2 from the ( x-3)/2 and then in antoehr line multiply with 2/2... and before u done that u could write sqrt( 1- (x-3)^2 /4) as sqrt(4 - ( x-3)^2)... ?
wait...at 5:07, why did (x-3)/2 turn into (x-3) SQUARED / 2?
nice one ^^
great!
You are loved by us BAŞKAAAAAAAAAN!
Also I swear you can't just put an inverse trig function inside a trig function so you would have to do the manipulations at the end anyway.
@jmlrugby11 scratch that you need 1/a^2 + u^2
could not agree more.
Ah well, it was an inadvertent error. He didn't carry it forward or anything.
You're a wizard!
made it so complicated at the end
why didn't he just use the arcsin integral? he made it way more complicated then it really is
he forgot to add plus c at the end!
plus C
you forgot + C at the end
nice. i had the same problem. the only difference is that mine is a definite integral. SMH. -_-
I made acouple typing errors. That identity should be
(cos(theta))^2 = 1 - (sin(theta))^2
zynot91210 sin2x is 2sinx cosx
You forgot the +C...
your hand hurts? what about my head, following you with the calculations. maple 15 reveals the same resolv.
+C or -1!
I love you Sal
Or just |Sin(θ)|.
I think he messed up, the calculus is fine but it should be 1- (x-3/4)^2, not 1- (x-3/2)^2. I don't think you can take the 4 out of the square root like that.
one does not simply correct Kahn
You can mate, look up your index laws.
He just used this identity: (a*b)^c = a^c*b^c
In this case, a is 4, b is 1-(x-3)^2/4, and c is 1/2
Therefore, (4*(1-(x-3)^2/4)^(1/2) is equivalent to 4^(1/2) times (1-(x-3)^2/4)^(1/2) which is equivalent to 2(1-(x-3)^2/4)^(1/2)
A square root is the same thing as taking the expression to the 1/2 power in case you didn't know.
I know this reply is too late, but it might help other people that think like you.
aw....man!!!
تحيا مصر
i seriously had a headache from following this...
Damn...
I think sin(2x)=sinxcosx+cosxsinx...
You're right, i noticed it too.
Cosmo Coralles same thing lol...u dumby
KiaTheMan how are they the same
KiaTheMan i underatand now
Your Ans. + C
Who doesn't?
holy shit.. i feel so stupid
this hurt my head ._.
applause
Answer is wrong. He forgot the +C... he wasted his whole video!
LMAO or as my professor would put it minus like 3 out of 8 points for sloppy notation.
WHERE IS YOUR +C :D
2theta - sin 2 theta + c, not +!!!! :)
nosebleeds
Lol+C ?
lol arsin
That's not the right answer... Sorry Charlie :)